When a drop of water is placed on a flat, clear surface such as a glass slide or plastic sheet, surface tension pulls the top surface into a curved, lens-like shape so that the drop functions as a simple magnifier. Suppose a drop of water has a maximum angular magnification of 3.50. (a) Find the drop’s focal length. (b) Assuming the bottom surface of the drop is flat, use the lens-maker’s equation from Topic 23 to calculate the radius of curvature of the top surface.

In a two-lens system, the focl length of the lens is f1 = 4.00 cm, and that of then lens 2 is f2 = 8.00 cm. The distance between the lenses is 16 cm. If an object O is 2.00 cm high and is placed 6.00 cm in front of the lens 1, where is the final image through the two-lens system?

Compute the focal length of a diverging thin lens made of flint glass, whose refractive index is 1.66 and is immersed in air having refractive index 1. The radii of the spherical surfaces of the lens are 10 cm and 20 cm. Select one: O -30 cm O O -10 cm 30 cm 10 cm

What is the focal length of the lens in your eye when you can focus on an object at your near point? Assume that the distance from your lens to the back of your eye is 2.0 cm, that your near point is 21 cm in front of your lens, and that the liquid behind the lens has n=1.00 just like the air in front of your lens. (a) 1.8 cm (b) 18 cm (c) 2.1 cm (d) 21 cm (e) None of the above.

It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance ss to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance s′s′ and then use the equation 1s+1s′=1f1s+1s′=1f, to calculate the focal length ff of the lens. But this procedure won'’t work with a diverging lens−−by itself, a diverging lens produces only virtual images, which can'’t be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object 20.0 cmcm to the left of the lens, the image is 29.7 cmcm to the right of the lens.…

The distance between the eyepiece and the objective lens in a certain compound microscope is 10.4 cm. The focal length of the objective is 0.430 cm, and that of the eyepiece is 1.40 cm. Find the overall magnification of the microscope. (The near point of the eye is 25 cm. Assume that the object is placed at the focal point of the objective lens, and one places the eyepiece at the near point of the eye.)

Suppose an object such as a book page is held 7.50 cm from a concave lens of focal length –10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. (a) Using the thin lens equations to calculate the location of the image. (b) What magnification is produced? (c) Use ray tracing to get an approximate location for the image.

A nearsighted person has a far point of 25.3cm and is prescribed contact lenses to correct her vision. What lens strength (a.k.a., lens power), in Diopters, should be prescribed? Give your answer using the usual convention of positive values for converging lenses and negative values for diverging lenses.